Area of Polygons

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SSAT Upper Level Quantitative › Area of Polygons

Questions 1 - 10

Find the area of a regular pentagon that has a side length of and an apothem of .

Explanation

To find the area of a regular polygon,

$\text{Area}=\frac{1}{2}(Perimeter)(apothem)$

To find the perimeter of the pentagon,

$\text{Perimeter}=5(side)$

For the given pentagon,

$\text{Perimeter}=5\times6=30$

So then, to find the area of the pentagon,

$\text{Area}=\frac{1}{2}(30)(2)=30$

Find the area of a regular hexagon that has side lengths of .

$54\sqrt3$

$36\sqrt3$

$48\sqrt3$

Explanation

Use the following formula to find the area of a regular hexagon:

$\text{Area}=\frac{3\sqrt3}{2}\times \text{side}^2$ .

Now, substitute in the length of the side into this equation.

$\text{Area}=\frac{3\sqrt3}{2}\times6^2$

$\text{Area}=\frac{3\sqrt3}{2}\times 36$

$\text{Area}=\frac{108\sqrt3}{2}=54\sqrt3$

The area of a rectangle is square feet. The width of the rectangle is four-sevenths of its length. Give the length of the rectangle in inches in terms of .

$6\sqrt{ 7K }$

$\frac{ \sqrt{ 7K } }{24}$

$\frac{ \sqrt{ 7K } }{2}$

$\frac{7K} {48}$

Explanation

Let be the length in feet. Then the width of the rectangle in feet is four-sevenths of this, or $\frac{4}{7}L$ . The area is equal to the product of the length and the width, so set up this equation and solve for :

$\frac{4}{7}L \cdot L = K$

$\frac{4}{7}L ^{2}= K$

$\frac{7} {4}\cdot \frac{4}{7}L ^{2}= \frac{7} {4}\cdot K$

$L ^{2}= \frac{7K} {4}$

$L =\sqrt{ \frac{7K} {4}} =\frac{ \sqrt{ 7K } }{\sqrt{4}}=\frac{ \sqrt{ 7K } }{2}$

Since this is the length in feet, we multiply this by 12 to get the length in inches:

$\frac{ \sqrt{ 7K } }{2} \cdot 12 = 6\sqrt{ 7K }$

A rectangle has the area of 80 square inches. The width of the rectangle is 2 inches longer that its height. Give the height of the rectangle.

Explanation

The area of a rectangle is given by multiplying the width times the height. That means:

where:

width and height.

We know that: $w=h+2\Rightarrow h=w-2$ . Substitube the in the area formula:

$Area=80=wh=(h+2)\times h\Rightarrow 80=h^2+2h\Rightarrow h^2+2h-80=0$

Now we should solve the equation for :

$h^2+2h-80=0\Rightarrow h=-1\pm \sqrt{1^2+1\times 80}=-1\pm \sqrt{81}=-1\pm 9$

The equation has two answers, one positive and one negative . As the length is always positive, the correct answer is inches.

Find the area of a regular pentagon that has a side length of and an apothem of .

Explanation

To find the area of a regular polygon,

$\text{Area}=\frac{1}{2}(Perimeter)(apothem)$

To find the perimeter of the pentagon,

$\text{Perimeter}=5(side)$

For the given pentagon,

$\text{Perimeter}=5\times6=30$

So then, to find the area of the pentagon,

$\text{Area}=\frac{1}{2}(30)(2)=30$

A rectangle has the area of 80 square inches. The width of the rectangle is 2 inches longer that its height. Give the height of the rectangle.

Explanation

The area of a rectangle is given by multiplying the width times the height. That means:

where:

width and height.

We know that: $w=h+2\Rightarrow h=w-2$ . Substitube the in the area formula:

$Area=80=wh=(h+2)\times h\Rightarrow 80=h^2+2h\Rightarrow h^2+2h-80=0$

Now we should solve the equation for :

$h^2+2h-80=0\Rightarrow h=-1\pm \sqrt{1^2+1\times 80}=-1\pm \sqrt{81}=-1\pm 9$

The equation has two answers, one positive and one negative . As the length is always positive, the correct answer is inches.

The area of a rectangle is square feet. The width of the rectangle is four-sevenths of its length. Give the length of the rectangle in inches in terms of .

$6\sqrt{ 7K }$

$\frac{ \sqrt{ 7K } }{24}$

$\frac{ \sqrt{ 7K } }{2}$

$\frac{7K} {48}$

Explanation

$\frac{4}{7}L \cdot L = K$

$\frac{4}{7}L ^{2}= K$

$\frac{7} {4}\cdot \frac{4}{7}L ^{2}= \frac{7} {4}\cdot K$

$L ^{2}= \frac{7K} {4}$

$L =\sqrt{ \frac{7K} {4}} =\frac{ \sqrt{ 7K } }{\sqrt{4}}=\frac{ \sqrt{ 7K } }{2}$

Since this is the length in feet, we multiply this by 12 to get the length in inches:

$\frac{ \sqrt{ 7K } }{2} \cdot 12 = 6\sqrt{ 7K }$

Find the area of a regular hexagon that has side lengths of .

$54\sqrt3$

$36\sqrt3$

$48\sqrt3$

Explanation

Use the following formula to find the area of a regular hexagon:

$\text{Area}=\frac{3\sqrt3}{2}\times \text{side}^2$ .

Now, substitute in the length of the side into this equation.

$\text{Area}=\frac{3\sqrt3}{2}\times6^2$

$\text{Area}=\frac{3\sqrt3}{2}\times 36$

$\text{Area}=\frac{108\sqrt3}{2}=54\sqrt3$

Parallelogram

Figure NOT drawn to scale.

The above depicts Rhombus , which has perimeter 80. .

Give the area of Rhombus .

Explanation

The area of any parallelogram is the product of the length of its base and that of a corresponding altitude. We can take $\overline{CD}$ as a base and perpendicular $\overline{XC}$ as an altitude.

All four sides of a rhombus have the same length, so we can find by dividing the perimeter - the sum of the lengths of the four sides - by 4:

$CD = p \div 4 = 80 \div 4 = 20$

Now multiply the lengths of this base and the altitude to get the area:

$a = CD \cdot XC = 20 \cdot 16 = 320$

Parallelogram

Figure NOT drawn to scale

The above figure shows Rhombus ; and are midpoints of their respective sides. Rhombus has area 900.

Give the area of Rectangle .

Explanation

A rhombus, by definition, has four sides of equal length. Therefore, , and, by the Multiplication Property, $\frac{1}{2 }AB=\frac{1}{2 } CD$ . Also, since and are the midpoints of their respective sides, $AX = XB = \frac{1}{2 }AB$ and $\frac{1}{2 } CD = CY = YD$ . Combining these statements, and letting :